//将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 
//
// 
//
// 示例 1： 
//
// 
//输入：l1 = [1,2,4], l2 = [1,3,4]
//输出：[1,1,2,3,4,4]
// 
//
// 示例 2： 
//
// 
//输入：l1 = [], l2 = []
//输出：[]
// 
//
// 示例 3： 
//
// 
//输入：l1 = [], l2 = [0]
//输出：[0]
// 
//
// 
//
// 提示： 
//
// 
// 两个链表的节点数目范围是 [0, 50] 
// -100 <= Node.val <= 100 
// l1 和 l2 均按 非递减顺序 排列 
// 
// Related Topics 递归 链表 
// 👍 2178 👎 0


//leetcode submit region begin(Prohibit modification and deletion)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 * int val;
 * ListNode next;
 * ListNode() {}
 * ListNode(int val) { this.val = val; }
 * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */

class Solution21 {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummyHead1 = new ListNode();
        dummyHead1.next = list1;
        ListNode dummyHead2 = new ListNode();
        dummyHead2.next = list2;

        ListNode resDummyHead = new ListNode();
        ListNode resTemp = resDummyHead;
        while (dummyHead1.next != null && dummyHead2.next != null) {
            if (dummyHead1.next.val <= dummyHead2.next.val) {
                resTemp.next = dummyHead1.next;
                resTemp = resTemp.next;
                dummyHead1.next = dummyHead1.next.next;
                resTemp.next = null;
            } else {
                resTemp.next = dummyHead2.next;
                resTemp = resTemp.next;
                dummyHead2.next = dummyHead2.next.next;
                resTemp.next = null;
            }
        }
        if (dummyHead1.next != null) {
            resTemp.next = dummyHead1.next;
        }
        if (dummyHead2.next != null) {
            resTemp.next = dummyHead2.next;
        }
        return resDummyHead.next;
    }
}
//leetcode submit region end(Prohibit modification and deletion)
